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Math Help - Help with multiplicatives

  1. #1
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    Help with multiplicatives

    Can anyone help with this question?

    Let f and g be multiplicative functions that are not identically zero and such that f(p^k)=g(p^k) for each prime p and k≥1. Prove that f = g.
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  2. #2
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    Quote Originally Posted by lyla View Post
    Can anyone help with this question?

    Let f and g be multiplicative functions that are not identically zero and such that f(p^k)=g(p^k) for each prime p and k≥1. Prove that f = g.
    We want to show that,

    f(n)=g(n) for any positive integer n.

    Your statement is not actually true . It is true if n>1.
    You need an addition fact that f(1)=g(1).
    But anyway, let me show that they agree on n>1.

    If n>1 we can prime factorize,
    n=p_1^(a_1) * p_2^(a_2) * ... * p_k^(a_k)

    Now,
    f(n)=f(p_1^(a_1)*p_2^(a_2)*...*p_k^(a_k))=f(p_1^(a _1))*...*f(p_k^(a_k)).
    Because,
    gcd(p_i,p_j)=1 if i!=j.

    The same thing for g(n).
    But,
    f(p_i^(a_i))=g(p_i^(a_i))
    Thus,
    f(n)=g(n) for n>1.
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  3. #3
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    Thank you very much, I was confused in seeing when they were equal.
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  4. #4
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    Quote Originally Posted by lyla View Post
    Thank you very much, I was confused in seeing when they were equal.
    Because,

    f(p_1^(a_1))=g(p_1^(a_1))
    f(p_2^(a_2))=g(p_2^(a_2))
    ...
    f(p_k^(a_k))=g(p_k^(a_k))

    Then multily them out,

    f(p_1^(a_1))*...*f(p_k^(a_k))=g(p_1^(a_1))*...*g(p _k^(a_k))
    Now you can put them under the same function because it is weakly multiplicative.

    f(p_1^(a_1)*...*p_k^(a_K))=g(p_1^(a_1)*...*p_k^(a_ k))

    But then,

    f(n)=g(n)
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